Question

Giúp mình câu 4 với

Answer 1

\(\left(cosx-sinx\right).sinx.cosx=cos.cos2x\)

\(\Leftrightarrow sinx.cos^2x-sin^2x.cosx=cos\left(1-2sin^2x\right)\)

\(\Leftrightarrow sinx.cos^2x=cosx-sin^2x.cosx\)

\(\Leftrightarrow sinx.cos^2x=cosx\left(1-sin^2x\right)\)

\(\Leftrightarrow sinx.cos^2x=cos^3x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sinx=cosx\end{matrix}\right.\)

Xét \(cos^2x=0\Leftrightarrow cosx=0\)\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

Xét \(sinx=cosx\) \(\Leftrightarrow sinx-cosx=0\) \(\Leftrightarrow\sqrt{2}.sin\left(x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)\(\left(k\in Z\right)\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)(\(k\in Z\))

Vậy \(x=\dfrac{\pi}{4}+k\pi\) hoặc \(x=\dfrac{\pi}{2}+k\pi\) với \(k\in Z\)